The ionic radii of K+, Na+, Al3+ and Mg2+ are in the order:

🧠 Three of these are isoelectronic, $\ce{K^+}$ is not — handle them in two steps $\ce{Na^+}$, $\ce{Mg^{2+}}$, $\ce{Al^{3+}}$ all have 10 electrons. $\ce{K^+}$ has 18 electrons (one extra shell). So you cannot rank all four by Z alone. First rank the three 10-electron ions, then place $\ce{K^+}$.

🗺️ Step 1 — the 10-electron trio Same electron cloud, ranked by Z: $\ce{Na^+}$ ($Z=11$) → biggest of the three. $\ce{Mg^{2+}}$ ($Z=12$). $\ce{Al^{3+}}$ ($Z=13$) → smallest.

So $\ce{Al^{3+}} < \ce{Mg^{2+}} < \ce{Na^+}$.

Step 2 — fit $\ce{K^+}$ into the order $\ce{K^+}$ has the configuration of Argon — one shell more than $\ce{Na^+}$. An extra shell beats any difference in nuclear pull at this level. So $\ce{K^+} > \ce{Na^+}$.

Final order: $\ce{Al^{3+}} < \ce{Mg^{2+}} < \ce{Na^+} < \ce{K^+}$.

⚠️ The trap Option (b) puts $\ce{K^+}$ between $\ce{Mg^{2+}}$ and $\ce{Na^+}$ as if all four were isoelectronic. They are not — $\ce{K^+}$ has 18 electrons. Always count electrons first. Mixing isoelectronic and non-isoelectronic ions in one question is a common JEE trap.

$\boxed{\text{Answer: (c)}}$